The radius of the base of a cone is decreasing at a rate of $2$ centimeters per minute. The height of the cone is fixed at $9$ centimeters. At a certain instant, the radius is $13$ centimeters. What is the rate of change of the volume of the cone at that instant (in cubic centimeters per minute)? Choose 1 answer: Choose 1 answer: (Choice A) A $-507\pi$ (Choice B) B $-156\pi$ (Choice C) C $-12\pi$ (Choice D) D $-78\pi$ The volume of a cone with radius $r$ and height $h$ is $\pi r^2\dfrac{h}{3}$.
Setting up the math Let... $r(t)$ denote the cone's radius at time $t$, and $V(t)$ denote the cone's volume at time $t$. We are given that $r'(t)=-2$ and that $r(t_0)=13$ for a specific time $t_0$. Note that $r'(t)$ is negative because the radius is decreasing. We are also given that the cone's height is $9$ centimeters. We want to find $V'(t_0)$. Relating the measures $V(t)$ and $r(t)$ relate to each other through the formula for the volume of a cone: $V(t)=\pi[r(t)]^2\dfrac{h}{3}$ Plugging $h=9$, we get the following equation: $V(t)=3\pi[r(t)]^2$ We can differentiate both sides to find an expression for $V'(t)$ : $V'(t)=6\pi r(t)r'(t)$ Using the information to solve Let's plug ${r(t_0)}={13}$ and ${r'(t_0)}={-2}$ into the expression for $V'(t_0)$ : $\begin{aligned} V'(t_0)&=6\pi{r(t_0)}{r'(t_0)} \\\\ &=6\pi({13})({-2}) \\\\ &=-156\pi \end{aligned}$ In conclusion, the rate of change of the volume of the cone at that instant is $-156\pi$ cubic centimeters per minute. Since the rate of change is negative, we know that the volume is decreasing.